Meta·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Meta phone screen for a SWE role, basically a warm-up question on implementing fast exponentiation. Pretty standard stuff but there are a couple of gotchas that can trip you up if you haven't drilled it recently.

Questions Asked (1)

Implement pow(x, n) where n can be negative, and achieve O(log |n|) time complexity.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The basic squaring logic isn't the hard part.

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Suggested Approach

Use binary exponentiation (exponentiation by squaring) to compute x^n in O(log |n|) time. Handle negative exponents by inverting x and negating n, and carefully manage edge cases like n=0, x=0, and integer overflow.

Pro tip: Discuss how to avoid integer overflow when negating n (e.g., if n is INT_MIN) and mention that using a long long for n or handling it separately shows attention to detail. Also, note that recursion uses O(log n) stack space, so an iterative approach is more space-efficient.

1. Clarify requirements and edge cases

Ask about constraints: Can x be 0? Can n be 0? What are the ranges of x and n? How to handle overflow? This shows thoroughness.

2. Explain the binary exponentiation algorithm

Describe how to compute x^n by repeatedly squaring x and halving n, using the binary representation of n. This achieves O(log |n|) time.

3. Handle negative exponents

If n is negative, compute pow(x, -n) and return 1/result. Be cautious with integer overflow when negating n, especially if n is INT_MIN.

4. Implement iteratively or recursively

Write clean code, either iteratively (using a loop) or recursively. Discuss trade-offs: recursion is simpler but uses O(log n) stack space; iteration is O(1) space.

5. Test with edge cases and analyze complexity

Test with n=0, negative n, x=0, and large values. Confirm time complexity O(log |n|) and space complexity O(1) for iterative.

Key Points to Mention

Binary exponentiation (exponentiation by squaring) reduces time complexity to O(log |n|).
Handling negative exponents by inverting the base and negating the exponent.
Edge cases: n=0 (result 1), x=0 (result 0 for positive n, undefined for negative n), and x=1 or x=-1.
Integer overflow when negating n (e.g., n = INT_MIN) and potential overflow in multiplication.
Space complexity: iterative approach uses O(1) space, recursive uses O(log n) due to call stack.
Using bitwise operations (n & 1, n >>= 1) for efficiency and clarity.

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